12. 如图,在$Rt△ ABC$中,$∠ ACB=90°$, D是边AB上一点,连结CD, $CD=CB$,过点A作CD的垂线,交CD的延长线于点E.求证:$△ ADE ∽ △ ABC.$

$\because D$ 是边$AB$ 上一点, $CD = CB, \therefore ∠ B = ∠ CDB.$ 又 $\because ∠ CDB = ∠ ADE, \therefore ∠ ADE = ∠ B. \because$ 过点 $A$ 作$CD$ 的垂线,交 $CD$ 的延长线于点 $E, \therefore AE ⊥ CE. \therefore ∠ E = 90°. \because ∠ ACB = 90°, \therefore ∠ E = ∠ ACB. \therefore △ ADE ∽ △ ABC$

13. 如图, 四边形 ABCD 是平行四边形, DE 交 BC 于点 F, 交 AB 的延长线于点 E, 且 $∠ EDB=∠ C$.
(1) 求证: $△ ADE ∽ △ DBE$;
(2) 若 $DC=9\ \mathrm{cm}$, $BE=16\ \mathrm{cm}$, 求 $DE$ 的长.

(第 13 题)

(1) $\because$ 四边形 $ABCD$ 是平行四边形, $\therefore ∠ A = ∠ C.$ 又 $\because ∠ EDB = ∠ C, \therefore ∠ EAD = ∠ EDB.$ 又 $\because ∠ AED = ∠ DEB, \therefore △ ADE ∽ △ DBE$
(2) $\because$ 四边形 $ABCD$ 是平行四边形, $DC = 9\ \mathrm{cm}, \therefore AB = DC = 9\ \mathrm{cm}.$ 又 $\because BE = 16\ \mathrm{cm}, \therefore AE = AB + BE = 9 + 16 = 25 (\mathrm{cm}).$ 由(1)可知, $△ ADE ∽ △ DBE. \therefore \frac{AE}{DE} = \frac{DE}{BE}, 即 \frac{25}{DE} = \frac{DE}{16}. \therefore DE = 20\ \mathrm{cm}$